Single Variable Calculus, Chapter 2, 2.4, Section 2.4, Problem 13

(a) Determine a number $\delta$ such that if $| x - 2 | < \delta$ then $| 4x - 8 | < \varepsilon$, where $\varepsilon = 0.1$

Based from the definition,

$\begin{equation} \begin{aligned} \begin{array}{c} \text{ if } |x -a| < \delta \text{ then } |f(x) - L| < \varepsilon\\ \text{ if } |x -2| < \delta \text{ then } |4x - 8| < 0.1 \end{array} \end{aligned} \end{equation}$


To satisfy inequatlity $| x - 2 | < \delta$

We want,

$\begin{equation} \begin{aligned} & | 4x - 8 | < 0 . 1 && \phantom{x}\\ & | 4(x - 2) | < 0.1 && \text{ Factor}\\ & \frac{4 | x - 2 |}{4} < \frac{0.1}{4} && \text{ Divide both sides by 4}\\ & | x- 2| < 0.025 && \end{aligned} \end{equation}$


Hence,
$\quad \delta < 0.025$

(b) Repeat part (a), where $\varepsilon = 0.01$

Using the definition,

$\begin{equation} \begin{aligned} & | 4x - 8 | < 0.01 && \\ & 4| x - 2 | < 0.001 && \text{ Factor }\\ & \frac{4| x-2|}{4} < \frac{0.01}{4} && \text{ Divide both sides by 4}\\ & |x-2| < 0.0025 && \\ \end{aligned} \end{equation}$

Hence,
$\quad \delta < 0.0025$

This means that by keeping $x$ within $0.0025$ of $2$, we are able to keep $f(x)$ within $0.1$ of $8$.

Although we chose $\delta = 0.0025$, any smaller positive value of $\delta$ would also have work.